Problem: The equation of a circle $C$ is $x^2+y^2-14x-2y+46 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Solution: To find the equation in standard form, complete the square. $(x^2-14x) + (y^2-2y) = -46$ $(x^2-14x+49) + (y^2-2y+1) = -46 + 49 + 1$ $(x-7)^{2} + (y-1)^{2} = 4 = 2^2$ Thus, $(h, k) = (7, 1)$ and $r = 2$.